Half-Line Events and Complements

Introduction

A CDF records probabilities up to and including a threshold. That phrase—up to and including—matters: if a random variable can equal the threshold with positive probability, changing a weak inequality to a strict one changes the answer. The complement gives the probability on the other side, but it changes the endpoint as well.

This tutorial organizes those endpoint choices for half-line events. The central distinction is between the CDF value \(F(a)\) and its left limit \(F(a-)\). Both are needed to calculate probabilities at a finite threshold \(a\); neither should be substituted for the other without checking whether there is probability concentrated at that point.

Four Half-Line Events at a Finite Threshold

Let \(X\) be a real-valued random variable with CDF \(F(x)=P(X\le x)\). For a finite real number \(a\), the four basic events are \(X\le a\), \(X<a\), \(X>a\), and \(X\ge a\). The CDF directly gives the first probability. The others follow either from the CDF's left limit or from taking a complement.

The two complement pairs are \(X\le a\) versus \(X>a\), and \(X<a\) versus \(X\ge a\). Notice that the endpoint belongs to exactly one event in each pair. Thus, subtracting \(F(a)\) from one gives the probability of being strictly above \(a\), whereas subtracting \(F(a-)\) from one gives the probability of being at or above \(a\).

Why the Left Limit Gives the Strict Inequality

The CDF is right-continuous, but the probability of a strict inequality looks to the left of the threshold. One way to see this precisely is to approach \(a\) using thresholds \(a-1/n\). The events \(\{X\le a-1/n\}\) increase as \(n\) increases, and their union is exactly \(\{X<a\}\).

Proof. Set \(A_n=\{X\le a-1/n\}\). Since \(a-1/n\) increases to \(a\), the events \(A_n\) increase. If \(X<a\), then \(a-X>0\), so there is an integer \(n\) large enough that \(1/n\le a-X\). Hence \(X\le a-1/n\), and \(X\) belongs to some \(A_n\). Conversely, every value in any \(A_n\) is strictly less than \(a\). Thus \(\bigcup_{n=1}^{\infty} A_n=\{X<a\}\).

For an increasing sequence of events, countable additivity implies continuity from below: the disjoint sets \(A_1\) and \(A_n\setminus A_{n-1}\) for \(n\ge2\) have union \(\bigcup_n A_n\), so the probability of that union is the limit of \(P(A_n)\). Therefore \(P(X<a)=\lim_{n\to\infty}P(A_n)=\lim_{n\to\infty}F(a-1/n)=F(a-)\). Taking the complement of \(\{X<a\}\) gives \(P(X\ge a)=1-F(a-)\). \(\square\)

The two other formulas follow directly from the definition of the CDF and the complement rule. The event \(\{X\le a\}\) has probability \(F(a)\), and its complement is \(\{X>a\}\), so \(P(X>a)=1-F(a)\). Also, \(\{X<a\}\) and \(\{X=a\}\) together make up \(\{X\le a\}\); equivalently, the jump from \(F(a-)\) to \(F(a)\) is the point probability: \(P(X=a)=F(a)-F(a-)\). This last identity helps explain exactly why the formulas for strict and weak inequalities can differ.

Atoms and Endpoint Decisions

A point \(a\) is an atom of the distribution when \(P(X=a)>0\). At such a point, the CDF jumps: \(F(a)>F(a-)\). The size of the jump is precisely the probability at the endpoint. In the absence of an atom, the CDF has no jump there and the strict and weak versions of a threshold probability agree.

If the distribution has no point mass at \(a\), then \(F(a)=F(a-)\), and consequently \(P(X<a)=P(X\le a)\) and \(P(X>a)=P(X\ge a)\). This is why endpoint distinctions can appear to disappear in continuous examples. But the absence of an atom must be justified by the distribution; it is not a property of CDF notation itself.

A Reliable Calculation Strategy

For a half-line probability, first write the event with its inequality sign intact. Then match that sign to the correct CDF value or complement. This prevents a common shortcut—using \(1-F(a)\) for every event that sounds like it is “above” \(a\)—from silently dropping an endpoint that belongs in the event.

Common Mistake

A frequent error is to reason that “at least \(a\)” means “one minus the probability up to \(a\),” and therefore write \(P(X\ge a)=1-F(a)\). But \(F(a)=P(X\le a)\), so \(1-F(a)=P(X>a)\): it excludes the endpoint. The correction is \(P(X\ge a)=1-F(a-)\), since \(F(a-)=P(X<a)\). In the discrete example, this distinction gives \(P(X\ge3)=1/2\), while \(1-F(3)=P(X>3)=1/3\). The difference, \(1/6\), is the mass at 3.